How do you use partial fractions to find the integral $\int \frac{x}{16 x^{4} - 1} d x$ $int x/(16x^4-1)dx$ ?

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How do you use partial fractions to find the integral $\int \frac{x}{16 x^{4} - 1} d x$ $int x/(16x^4-1)dx$ ?

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Answer 1 · 2017-07-24T09:35:40

The answer is $= \frac{1}{16} ln (| 2 x + 1 |) + \frac{1}{16} ln (| 2 x - 1 |) - \frac{1}{16} ln (∣ ∣ 4 x^{2} + 1 ∣ ∣) + C$ $=1/16ln(|2x+1|)+1/16ln(|2x-1|)-1/16ln(|4x^2+1|)+C$

Explanation:

We need

$\int \frac{d x}{x} = ln (| x |) + C$ $intdx/x=ln(|x|)+C$

Let's factorise the denominator

$16 x^{4} - 1 = (4 x^{2} - 1) (4 x^{2} + 1) = (2 x + 1) (2 x - 1) (4 x^{2} + 1)$ $16x^4-1=(4x^2-1)(4x^2+1)=(2x+1)(2x-1)(4x^2+1)$

We can perform the decomposition into partial fractions

$\frac{x}{16 x^{4} - 1} = \frac{x}{(2 x + 1) (2 x - 1) (4 x^{2} + 1)}$ $x/(16x^4-1)=x/((2x+1)(2x-1)(4x^2+1))$

$= \frac{A}{2 x + 1} + \frac{B}{2 x - 1} + \frac{C x + D}{4 x^{2} + 1}$ $=A/(2x+1)+B/(2x-1)+(Cx+D)/(4x^2+1)$

$= \frac{A (2 x - 1) (4 x^{2} + 1) + B (2 x + 1) (4 x^{2} + 1) + (C x + D) (2 x + 1) (2 x - 1)}{(2 x + 1) (2 x - 1) (4 x^{2} +)}$ $=(A(2x-1)(4x^2+1)+B(2x+1)(4x^2+1)+(Cx+D)(2x+1)(2x-1))/((2x+1)(2x-1)(4x^2+))$

The denominators are the same, we compare the numerators

$x = A (2 x - 1) (4 x^{2} + 1) + B (2 x + 1) (4 x^{2} + 1) + (C x + D) (2 x + 1) (2 x - 1)$ $x=A(2x-1)(4x^2+1)+B(2x+1)(4x^2+1)+(Cx+D)(2x+1)(2x-1)$

Let $x = - \frac{1}{2}$ $x=-1/2$ , $\Rightarrow$ $=>$ , $- \frac{1}{2} = - 4 A$ $-1/2=-4A$ , $\Rightarrow$ $=>$ , $A = \frac{1}{8}$ $A=1/8$

Let $x = \frac{1}{2}$ $x=1/2$ , $\Rightarrow$ $=>$ , $\frac{1}{2} = 4 B$ $1/2=4B$ , $\Rightarrow$ $=>$ , $B = \frac{1}{8}$ $B=1/8$

Let $x = 0$ $x=0$ , $\Rightarrow$ $=>$ , $0 = - A + B - D$ $0=-A+B-D$ , $\Rightarrow$ $=>$ , $D = 0$ $D=0$

Coefficients of $x^{3}$ $x^3$ ,

$0 = 8 A + 8 B + 4 C$ $0=8A+8B+4C$ , $\Rightarrow$ $=>$ , $4 C = - 2$ $4C=-2$ , $C = - \frac{1}{2}$ $C=-1/2$

Therefore,

$\frac{x}{16 x^{4} - 1} = \frac{\frac{1}{8}}{2 x + 1} + \frac{\frac{1}{8}}{2 x - 1} + \frac{- \frac{1}{2} x}{4 x^{2} + 1}$ $x/(16x^4-1)=(1/8)/(2x+1)+(1/8)/(2x-1)+(-1/2x)/(4x^2+1)$

$\int \frac{x d x}{16 x^{4} - 1} = \int \frac{\frac{1}{8} d x}{2 x + 1} + \int \frac{\frac{1}{8} d x}{2 x - 1} + \int \frac{- \frac{1}{2} x d x}{4 x^{2} + 1}$ $int(xdx)/(16x^4-1)=int(1/8dx)/(2x+1)+int(1/8dx)/(2x-1)+int(-1/2xdx)/(4x^2+1)$

$= \frac{1}{16} ln (| 2 x + 1 |) + \frac{1}{16} ln (| 2 x - 1 |) - \frac{1}{16} ln (∣ ∣ 4 x^{2} + 1 ∣ ∣) + C$ $=1/16ln(|2x+1|)+1/16ln(|2x-1|)-1/16ln(|4x^2+1|)+C$

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How do you use partial fractions to find the integral $\int \frac{x}{16 x^{4} - 1} d x$ $int x/(16x^4-1)dx$ ?

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